Come on guys it's not that hard

Okay, this is probably wrong. Answer is: $\left \lceil{\frac{m}{2}}\right \rceil+\left \lceil{\frac{n}{2}}\right \rceil$ if one of $m,n$ is even and $\left \lceil{\frac{m}{2}}\right \rceil+\left \lceil{\frac{n}{2}}\right \rceil-1$ otherwise. One can clearly see that each boundary column and row must contain at least $1$ token. We consider the bottom row , there is at least one token in this row $T_1$, if in the two above rows there is no token , we can never fullfill the table with tokens,so there is at least one token in the above two rows ,and since we are considering minimal number ,this token should not be in adjacent row with $T_1$ ,and if this token is not in the same column with $T_1$ we can't fullfill the table , so these two must be in the same column, continuing like this there must be at least $\left \lceil{\frac{n}{2}}\right \rceil$ tokens in this column . Doing the same for colums we can see that there are at least $\left \lceil{\frac{m}{2}}\right \rceil$ tokens in some row. And since a column and a row coincide in at most $1$ square , we must have at least $\left \lceil{\frac{m}{2}}\right \rceil+\left \lceil{\frac{n}{2}}\right \rceil-1$ tokens. For $n,m$ both odd we can give an example of $\left \lceil{\frac{m}{2}}\right \rceil+\left \lceil{\frac{n}{2}}\right \rceil-1$ tokens , let one token in top right corner ,and start putting tokens in the first row and first column, such that each token is put in a square 2 column/row away from the last token put in that column/row, clearly this works. If W.LO.G $n$ even , a column has at least $\left \lceil{\frac{n}{2}}\right \rceil$ tokens , and leaves one boundary square without token , now if we don't put an extra token in this square then the column with $\left \lceil{\frac{m}{2}}\right \rceil$ has to go through this boundary square ,so we can't do it with $\left \lceil{\frac{m}{2}}\right \rceil+\left \lceil{\frac{n}{2}}\right \rceil-1$ tokens, and with $\left \lceil{\frac{m}{2}}\right \rceil+\left \lceil{\frac{n}{2}}\right \rceil$ tokens we can give a similar with above construction.

The answer is $\lceil \frac{m+n}{2} \rceil$. The construction is pretty easy. To show that $M \ge \lceil \frac{m+n}{2} \rceil$.Consider the total perimeter of the shapes created by the blackened cells (namely the number of black and white neighbouring cells). This total perimeter is a monoinvariant,apparently our move does not increase it. So the perimeter at the beginning is $4M$ and at the end it should be $2m+2n$ thus $4M \ge 2(m+n)$.

very nice! my friend

mela_20-15 wrote: The answer is $\lceil \frac{m+n}{2} \rceil$. The construction is pretty easy. To show that $M \ge \lceil \frac{m+n}{2} \rceil$.Consider the total perimeter of the shapes created by the blackened cells (namely the number of black and white neighbouring cells). This total perimeter is a monoinvariant,apparently our move does not increase it. So the perimeter at the beginning is $4M$ and at the end it should be $2m+2n$ thus $4M \ge 2(m+n)$. How do you know there is an example for $M$$=$$\lceil$$\frac{m+n}{2}$$\rceil$